A 1600.0 g sample of ice at 0.0 deg. C is heated until it is liquid water at 42

Question

A 1600.0 g sample of ice at 0.0 deg. C is heated until it is liquid water at 42 dec.C How much energy was added to the sample?

Material csolid Tm Lf cliquid Tb Lv cg k x 10-6 x 10-6 x 103 x 103 K-1 K-1 J kg-1 K-1 oC J kg-1 J kg-1 K-1 oC J kg-1 J kg-1 K-1 J s-1 m-1 K-1 Al 26 75 900 660 380 900 2450 11400 900 240 Ammonia -- -- 4700 -77.8 3320 4700 -33.4 1371 2060 -- Brass 19 56 -- 920 -- -- -- -- -- 110 Concrete 12 36 -- -- -- -- -- -- -- 1.1 Copper 17 51 378 1083 134 378 2565 5069 378 390 Ethanol -- 1100 2450 -114.4 1080 2450 78.3 855 1680 -- Gasoline -- 950 -- -- -- 2220 -- -- -- -- Glass 9 27 670 1500 -- -- -- -- -- 0.8 Glycol -- 570 -- -- -- 2360 -- -- -- -- Gold 14 42 129 1063 64.5 129 2660 1578 129 -- Goose Down -- -- -- -- -- -- -- -- -- 0.025 Hydrogen -- -- 967 -259.3 58.6 967 -252.9 452 1432 -- Invar 0.9 2.7 460 1427 -- -- -- -- -- -- Iron 12 35 449 1538 -- -- 2862 -- -- 79 Kerosene -- 990 -- -- -- 2010 -- -- -- -- Lead 29 87 128 327 24.5 128 1750 871 128 35 Nitrogen -- -- 2042 -210 25.5 2042 -195.8 201 1040 -- Oxygen -- -- 1669 -218.8 13.8 1669 -183 213 919 -- Pyrex 3 9 753 820 -- -- -- -- -- -- Silver 18 75 233 1763 -- -- 2162 -- -- 420 Steel 12 35 449 -- -- -- -- -- -- 14 Styrofoam -- -- -- -- -- -- -- -- -- 0.01 Water -- 214 211 0 334 4186 100 2256 2080 0.6 Wood -- -- -- -- -- -- -- -- -- 0.15 Wool -- -- -- -- -- -- -- -- -- 0.04

Explanation / Answer

The amount of heat requred to raise the temperature of ice from 0 degrre to 42 degree C

Q = mLf + mc delT

=1.6 kg ( 334 * 10^3 J/kg ) + 1.6 ( 4186) ( 42-0)

=815699.2 J

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